3.3: Solving Quadratic Equations of the Form- (x-a)²=c
- Page ID
- 20037
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)When solving equations of the form: \({\left(x-a\right)}^2=c\), we can use the square root property which states to drop the square and take the positive and negative square root of the constant, c. Mathematically, we say the following:
If \({\left(x-a\right)}^2=c\),
then \(x-a=\pm \sqrt{c}\)
next, add a on both sides of the equation to obtain
\(x=a\pm \sqrt{c}\)
Solve \({\left(x-5\right)}^2=16\)
Solution
To solve, we first drop the square from the left side and take the positive and negative square root of 16 to obtain
\[x-5=\pm \sqrt{16}\]
Next, we simplify the square root
\[x-5=\pm 4\]
Next, we isolate the variable, x by adding 5 on both sides of the equation:
\[x-5+\boldsymbol{5}=+\boldsymbol{5}\pm 4\]
Such that after simplifying, we obtain two answers
\[x=5\pm 4\]
\[x=5+4 \,\,\text{ and }\,\, x=5-4\]
\[x=9\,\, \text{ and }\,\, x=1\]
Solve \(3{\left(x+2\right)}^2+5=80\)
Solution
To solve, we must first isolate the perfect square to create the form: \({\left(x-a\right)}^2=c\), so we first subtract 5 on both sides of the equation, then divide both sides by 3
\[3{\left(x+2\right)}^2+5-\boldsymbol{5}=80-\boldsymbol{5}\]
\[3{\left(x+2\right)}^2=75\]
\[\frac{3\left(x+2\right)^2}{\boldsymbol{3}}=\frac{75}{\boldsymbol{3}}\]
\[{\left(x+2\right)}^2=25\]
Next, we drop the square and take the positive and negative square root of 25 to obtain
\[x+2=\pm \sqrt{25}\]
\[x+2=\pm 5\]
Next, we isolate the x by subtracting 2 on both sides of the equation
\[x+2-\boldsymbol{2}=-\boldsymbol{2}\pm 5\]
After simplifying, we obtain two answers
\[x=-2\pm 5\]
\[x=-2+5 \,\,\text{ and }\,\, x=-2-5\]
\[x=3 \,\,\text{ and }\,\, x=-7\]